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8/13/2019 Ch 10 Lecture
1/39
Chapter 10 - 1
ISSUES TO ADDRESS... When we combine two elements...
what is the resulting equilibrium state?
In particular, if we specify...
-- the composition (e.g., wt% Cu - wt% Ni), and
-- the temperature (T
)then...
How many phases form?
What is the composition of each phase?
What is the amount of each phase?
Chapter 10: Phase Diagrams
Phase BPhase A
Nickel atom
Copper atom
8/13/2019 Ch 10 Lecture
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Chapter 10 - 2
Phase Equilibria: Solubility Limit
Question: What is the
solubility limit for sugar inwater at 20C?
Answer: 65 wt% sugar.At 20C, if C< 65 wt% sugar:syrup
At 20C,if C> 65 wt% sugar:syrup + sugar
65
Solubility Limit:Maximum concentration forwhich only a single phase
solution exists.
Sugar/Water Phase Diagram
S
ugar
Temperature(C)
0 20 40 60 80 100C= Composition (wt% sugar)
L (liquid solution
i.e., syrup)
SolubilityLimit L
(liquid)
+S
(solidsugar)20
406080
100
W
ater
Adapted from Fig. 10.1,
Callister & Rethwisch 4e.
Solutionsolid, liquid, or gas solutions, single phase Mixturemore than one phase
8/13/2019 Ch 10 Lecture
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Chapter 10 - 3
Components:
The elements or compounds which are present in the alloy(e.g., Al and Cu)
Phases:The physically and chemically distinct material regions
that form (e.g., aand b).
Aluminum-
Copper
Alloy
Components and Phases
a (darkerphase)
b(lighter
phase)
Adapted from chapter-
opening photograph,
Chapter 9, Callister,
Materials Science &
Engineering: An
Introduction, 3e.
8/13/2019 Ch 10 Lecture
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Chapter 10 - 4
70 80 1006040200
Temperature(C
)
C = Composition (wt% sugar)
L(liquid solution
i.e., syrup)
20
100
40
60
80
0
L(liquid)+S
(solidsugar)
Effect of Temperature & Composition
Altering Tcan change # of phases: pathAto B.
Altering Ccan change # of phases: path Bto D.
water-
sugar
system
Adapted from Fig. 10.1,
Callister & Rethwisch 4e.
D(100C,C= 90)2 phases
B(100C,C= 70)1 phase
A(20C,C= 70)2 phases
8/13/2019 Ch 10 Lecture
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8/13/2019 Ch 10 Lecture
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Chapter 10 - 6
Phase Diagrams
Indicate phases as a function of T, C, and P.
For this course:- binary systems: just 2 components.
- independent variables: Tand C (P= 1 atm is almost always used).
PhaseDiagram
for Cu-Ni
system
Adapted from Fig. 10.3(a), Callister &
Rethwisch 4e.(Fig. 10.3(a) is adapted
from Phase Diagrams of Binary Nickel
Alloys, P. Nash (Ed.), ASM International,
Materials Park, OH (1991).
2 phases:L(liquid)a(FCC solid solution)
3 different phase fields:LL+
a
a
wt% Ni20 40 60 80 10001000110012001300140015001600T(C)
L(liquid)
a(FCC solidsolution)
8/13/2019 Ch 10 Lecture
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Chapter 10 - 7
Cu-Ni
phase
diagram
Isomorphous Binary Phase Diagram
Phase diagram:
Cu-Ni system.
System is:
Adapted from Fig. 10.3(a), Callister &
Rethwisch 4e.(Fig. 10.3(a) is adapted
from Phase Diagrams of Binary Nickel
Alloys, P. Nash (Ed.), ASM International,
Materials Park, OH (1991).
-- binaryi.e., 2 components:
Cu and Ni.
-- isomorphousi.e., complete
solubility of one
component in
another; aphase
field extends from
0 to 100 wt% Ni.wt% Ni20 40 60 80 10001000
110012001300140015001600T(C)
L(liquid)
a(FCC solidsolution)
8/13/2019 Ch 10 Lecture
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Chapter 10 -
wt% Ni20 40 60 80 1000100011001200
1300140015001600T(C)
L(liquid)
a(FCC solid
solution)
Cu-Ni
phase
diagram
8
Phase Diagrams:Determination of phase(s) present
Rule 1: If we know Tand Co, then we know:-- which phase(s) is (are) present.
Examples:
A(1100C, 60 wt% Ni):
1 phase: a
B(1250C, 35 wt% Ni):2 phases: L+ a
B(12
50
C,35)
A (1100C,60)Adapted from Fig. 10.3(a), Callister &Rethwisch 4e.(Fig. 10.3(a) is adaptedfrom Phase Diagrams of Binary Nickel
Alloys, P. Nash (Ed.), ASM International,
Materials Park, OH (1991).
8/13/2019 Ch 10 Lecture
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Chapter 10 - 9
wt% Ni201200
1300
T(C)L(liquid)
a(solid)
30 40 50
Cu-Ni
system
Phase Diagrams:Determination of phase compositions
Rule 2: If we know Tand C0, then we can determine:-- the composition of each phase.
Examples:TA A
35C032CL
At TA = 1320C:Only Liquid (L) presentCL= C0 ( = 35 wt% Ni)
At TB = 1250C:Both aand L presentCL = Cliquidus ( = 32 wt% Ni)Ca = Csolidus ( = 43 wt% Ni)
At TD = 1190C:Only Solid (a) presentCa= C0 ( = 35 wt% Ni)
Consider C0= 35 wt% Ni
DTD
tie line
4Ca3Adapted from Fig. 10.3(a), Callister &
Rethwisch 4e.(Fig. 10.3(a) is adapted from
Phase Diagrams of Binary Nickel Alloys, P.
Nash (Ed.), ASM International, Materials
Park, OH (1991).
BTB
8/13/2019 Ch 10 Lecture
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Chapter 10 - 10
Rule 3: If we know Tand C0, then can determine:-- the weight fraction of each phase. Examples:
At TA : Only Liquid (L) present
WL = 1.00, Wa= 0
At TD : Only Solid (a) presentWL = 0, Wa = 1.00
Phase Diagrams:Determination of phase weight fractions
wt% Ni201200
1300
T(C)L(liquid)
a(solid)
30 40 50
Cu-Ni
system
TAA
35C0
32CL
BTB
DTD
tie line
4Ca3
R SAt TB : Both a and L present
73.03243
3543=
=
= 0.27
WL= SR+SWa= RR+S
Consider C0= 35 wt% Ni
Adapted from Fig. 10.3(a), Callister &
Rethwisch 4e.(Fig. 10.3(a) is adapted from
Phase Diagrams of Binary Nickel Alloys, P.
Nash (Ed.), ASM International, Materials
Park, OH (1991).
8/13/2019 Ch 10 Lecture
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Chapter 10 - 11
Tie lineconnects the phases in equilibrium with
each otheralso sometimes called an isotherm
The Lever Rule
What fraction of each phase?
Think of the tie line as a lever
(teeter-totter)
ML Ma
R S
M x S=MLx R
L
L
LL
L
LCC
CC
SR
RW
CC
CC
SR
S
MM
MW
=
+
=
=
+
=
+
=
a
a
a
a
a
00
wt% Ni20
1200
1300T(C)
L(liquid)
a(solid)
30 40 50
BTB
tie line
C0CL CaSR
Adapted from Fig. 10.3(b),
Callister & Rethwisch 4e.
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8/13/2019 Ch 10 Lecture
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Chapter 10 -
Slow rate of cooling:Equilibrium structure
Fast rate of cooling:Cored structure
First ato solidify:46 wt% Ni
Last ato solidify:
< 35 wt% Ni
13
Cachanges as we solidify.
Cu-Ni case: First ato solidify has Ca= 46 wt% Ni.Last ato solidify has Ca= 35 wt% Ni.
Cored vs Equilibrium Structures
Uniform Ca:35 wt% Ni
8/13/2019 Ch 10 Lecture
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Chapter 10 - 14
Mechanical Properties:Cu-Ni System
Effect of solid solution strengthening on:
-- Tensile strength (TS) -- Ductility (%EL)
Adapted from Fig. 10.6(a),
Callister & Rethwisch 4e.
TensileStrength
(MPa)
Composition, wt% NiCu Ni
0 20 40 60 80 100200
300
400
TSforpure Ni
TSfor pure Cu
Elongation
(%EL)
Composition, wt% NiCu Ni
0 20 40 60 80 10020
30
40
50
60
%ELfor
pure Ni
%ELfor pure Cu
Adapted from Fig. 10.6(b),
Callister & Rethwisch 4e.
8/13/2019 Ch 10 Lecture
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Chapter 10 - 15
2 components
has a special composition
with a min. melting T.
Adapted from Fig. 10.7,
Callister & Rethwisch 4e.
Binary-Eutectic Systems
3 single phase regions
(L, a, b)
Limited solubility:a: mostly Cu
b: mostly Ag
TE : No liquid below TE
: Composition at
temperature TE
CE
Ex.: Cu-Ag system
Cu-Ag
system
L(liquid)
a L+ a L+bba + b
C, wt% Ag20 40 60 80 1000
200
1200T(C)
400
600
800
1000
CE
TE 8.0 71.9 91.2779C
Ag)wt%1.29(Ag)wt%.08(Ag)wt%9.71( b+L
cooling
heating
Eutectic reaction
L(CE) a(CaE) + b(CbE)
8/13/2019 Ch 10 Lecture
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Chapter 10 - 16
L+a L+b
a+b
200
T(C)
18.3
C, wt% Sn
20 60 80 1000
300
100
L(liquid)
a 183C61.9 97.8
b
For a 40 wt% Sn-60 wt% Pb alloy at 150C, determine:
-- the phases present Pb-Snsystem
EX 1: Pb-Sn Eutectic System
Answer:a+ b-- the phase compositions
-- the relative amountof each phase
150
40C0
11Ca
99Cb
SR
Answer:Ca= 11 wt% SnCb= 99 wt% Sn
Wa= Cb- C0Cb- Ca=
99 - 4099 - 11
=5988
= 0.67
SR+S
=
Wb=C0 - CaCb - Ca
=R
R+S
=29
88= 0.33=
40 - 11
99 - 11
Answer:
Adapted from Fig. 10.8,
Callister & Rethwisch 4e.
8/13/2019 Ch 10 Lecture
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Chapter 10 - 17
Answer:Ca= 17 wt% Sn
-- the phase compositions
L+b
a+b
200
T(C)
C, wt% Sn
20 60 80 1000
300
100
L(liquid)
a bL+a
183C
For a 40 wt% Sn-60 wt% Pb alloy at 220C, determine:
-- the phases present: Pb-Snsystem
EX 2: Pb-Sn Eutectic System
-- the relative amountof each phase
Wa= CL- C0CL- Ca
=46 - 40
46 - 17
=6
29= 0.21
WL=C0- CaCL- Ca
=23
29= 0.79
40C0
46CL
17Ca
220 SR
Answer:a+ L
CL= 46 wt% Sn
Answer:
Adapted from Fig. 10.8,
Callister & Rethwisch 4e.
8/13/2019 Ch 10 Lecture
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Chapter 10 - 18
For alloys for which
C0< 2 wt% Sn
Result: at room temperature-- polycrystalline with grains of
aphase havingcomposition C0
Microstructural Developments
in Eutectic Systems I
0
L+a200
T(C)
C , wt% Sn10
220
C0
300
100
L
a
30
a+b
400
(room Tsolubility limit)
TE(Pb-SnSystem)
aLL: C0wt% Sn
a: C0wt% Sn
Adapted from Fig. 10.11,
Callister & Rethwisch 4e.
8/13/2019 Ch 10 Lecture
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Chapter 10 - 19
For alloys for which
2 wt% Sn < C0< 18.3 wt% Sn
Result:
at temperatures in a+ brange
-- polycrystalline with agrains
and small b-phaseparticles
Adapted from Fig. 10.12,
Callister & Rethwisch 4e.
Microstructural Developments
in Eutectic Systems II
Pb-Snsystem
L+ a200
T(C)
C, wt% Sn10
18.3
200C0
300
100
L
a
30
a+b
400
(sol. limit at TE)
TE
2(sol. limit at Troom)
L
a
L: C0wt% Sn
ab
a: C0wt% Sn
8/13/2019 Ch 10 Lecture
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Chapter 10 - 20
For alloy of composition C0= C
E
Result: Eutectic microstructure (lamellar structure)-- alternating layers (lamellae) of aand bphases.
Adapted from Fig. 10.13,Callister & Rethwisch 4e.
Microstructural Developments
in Eutectic Systems III
Adapted from Fig. 10.14,
Callister & Rethwisch 4e.
160m
Micrograph of Pb-Sneutectic
microstructure
Pb-Sn
system
L+b
a + b
200
T(C)
C, wt% Sn
20 60 80 1000
300
100
L
a bL+a
183C
40
TE
18.3
a: 18.3 wt%Sn
97.8
b: 97.8 wt% Sn
CE
61.9
L: C0wt% Sn
8/13/2019 Ch 10 Lecture
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Chapter 10 - 21
Lamellar Eutectic Structure
Adapted from Figs. 10.14 & 10.15,
Callister & Rethwisch 4e.
8/13/2019 Ch 10 Lecture
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Chapter 10 - 22
For alloys for which 18.3 wt% Sn < C0< 61.9 wt% Sn Result: aphase particles and a eutectic microconstituent
Microstructural Developments
in Eutectic Systems IV
18.3 61.9
SR
97.8
SR
primary aeutectic a
eutectic b
WL= (1-Wa)= 0.50
Ca = 18.3 wt% Sn
CL = 61.9 wt% SnS
R+ SWa= = 0.50
Just above TE :
Just below TE:
Ca = 18.3 wt% SnCb = 97.8 wt% Sn
S
R+ SWa= = 0.73
Wb= 0.27Adapted from Fig. 10.16,
Callister & Rethwisch 4e.
Pb-Sn
systemL+
b200
T(C)
C, wt% Sn
20 60 80 1000
300
100
L
a bL+a
40
a+bTE
L: C0wt% Sn LaL
a
8/13/2019 Ch 10 Lecture
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Chapter 10 - 23
L+aL+b
a+b200
C, wt% Sn20 60 80 1000
300
100
L
a bTE
40
(Pb-SnSystem)
Hypoeutectic & Hypereutectic
Adapted from Fig. 10.8,
Callister & Rethwisch 4e.
(Fig. 10.8 adapted from
Binary Phase Diagrams,
2nd ed., Vol. 3, T.B.
Massalski (Editor-in-Chief),
ASM International,
Materials Park, OH, 1990.)
160 m
eutectic micro-constituentAdapted from Fig. 10.14,
Callister & Rethwisch 4e.
hypereutectic: (illustration only)
bbb
bbb
Adapted from Fig. 10.17,
Callister & Rethwisch 4e.
(Illustration only)
(Figs. 10.14 and
10.17 from Metals
Handbook, 9th ed.,
Vol. 9,
Metallography and
Microstructures,
American Society for
Metals, Materials
Park, OH, 1985.)
175 m
aaa
aaa
hypoeutectic: C0= 50 wt% Sn
Adapted from
Fig. 10.17, Callister &
Rethwisch 4e.
T(C)
61.9
eutectic
eutectic: C0=61.9wt% Sn
8/13/2019 Ch 10 Lecture
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Chapter 10 - 24
Intermetallic Compounds
Mg2Pb
Note: intermetallic compound exists as a line on the diagram - not an
area - because of stoichiometry (i.e. composition of a compound
is a fixed value).
Adapted from
Fig. 10.20, Callister &
Rethwisch 4e.
8/13/2019 Ch 10 Lecture
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Chapter 10 - 25
Eutectoidone solid phase transforms to two other
solid phases
S2 S1+S3
a+ Fe3C (For Fe-C, 727C, 0.76 wt% C)
intermetallic compound
- cementite
cool
heat
Eutectic, Eutectoid, & Peritectic
Eutectic- liquid transforms to two solid phases
L a+ b (For Pb-Sn, 183C, 61.9 wt% Sn)coolheat
cool
heat
Peritectic- liquid and one solid phase transform to a
second solid phaseS1 + L S2
+ L (For Fe-C, 1493C, 0.16 wt% C)
8/13/2019 Ch 10 Lecture
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Chapter 10 - 26
Eutectoid & Peritectic
Cu-Zn Phase diagram
Adapted from Fig. 10.21,
Callister & Rethwisch 4e.
Eutectoid transformation +
Peritectic transformation + L
8/13/2019 Ch 10 Lecture
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Chapter 10 - 27
Ceramic Phase Diagrams
MgO-Al2O3 diagram:
Adapted from Fig.
10.24, Callister &
Rethwisch 4e.
8/13/2019 Ch 10 Lecture
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Chapter 10 - 28
Iron-Carbon (Fe-C) Phase Diagram
2 important
points
- Eutectoid (B):
a+Fe3C
- Eutectic (A):L +Fe3C
Adapted from Fig. 10.28,Callister & Rethwisch 4e.
Fe3C(cementite)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3C
a+Fe3C
(Fe) C, wt% C
1148C
T(C)
a 727C = Teutectoid
4.30
Result: Pearlite =alternating layers ofaand Fe3C phases
120 m
(Adapted from Fig. 10.31,Callister & Rethwisch 4e.)
0.76
B
AL+Fe3C
Fe3C (cementite-hard)
a(ferrite-soft)
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8/13/2019 Ch 10 Lecture
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Chapter 10 - 30
Fe3C(cem
entite)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3C
a+Fe3C
L+Fe3C
(Fe) C, wt% C
1148C
T(C)
a727C
(Fe-C
System)
C0
0.76
Hypoeutectoid Steel
a
aa
srWa= s/(r+s)W=(1 - Wa)
RSa
pearlite
Wpearlite=W
Wa= S/(R+S)
W =(1Wa)Fe3C
Adapted from Figs. 10.28
and 10.33,Callister &Rethwisch 4e.
(Fig. 10.28 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-in-
Chief), ASM International,
Materials Park, OH,
1990.)
Adapted from Fig. 10.34, Callister & Rethwisch 4e.
proeutectoid ferritepearlite
100 mHypoeutectoid
steel
8/13/2019 Ch 10 Lecture
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Chapter 10 - 31
Hypereutectoid Steel
Fe3C(cem
entite)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3C
a+Fe3C
L+Fe3C
(Fe) C, wt%C
1148C
T(C)
a
Adapted from Figs. 10.28
and 10.36,Callister &Rethwisch 4e. (Fig.
10.28 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-in-
Chief), ASM International,
Materials Park, OH,
1990.)
(Fe-C
System)
0.7
6 C0
Fe3C
Adapted from Fig. 10.37, Callister & Rethwisch 4e.
proeutectoid Fe3C
60 mHypereutectoidsteel
pearlite
pearlite
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Chapter 10 - 32
Fe3C(cem
entite)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3C
a+Fe3C
L+Fe3C
(Fe) C, wt%C
1148C
T(C)
a
Hypereutectoid Steel
(Fe-C
System)
0.7
6 C0
pearlite
Fe3C
xv
V X
Wpearlite=W
Wa=X/(V+X)
W =(1 - Wa)Fe3C
W =(1-W)
W =x/(v+x)
Fe3C
Adapted from Fig. 10.37, Callister & Rethwisch 4e.
proeutectoid Fe3C
60 mHypereutectoidsteel
pearlite
Adapted from Figs. 10.28
and 10.36,Callister &Rethwisch 4e. (Fig.
10.28 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-in-
Chief), ASM International,
Materials Park, OH,
1990.)
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Chapter 10 - 33
Example Problem
For a 99.6 wt% Fe-0.40 wt% C steel at atemperature just below the eutectoid,
determine the following:
a) The compositions of Fe3C and ferrite (a).
b) The amount of cementite (in grams) thatforms in 100 g of steel.
c) The amounts of pearlite and proeutectoid
ferrite (a) in the 100 g.
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Chapter 10 - 34
Solution to Example Problem
WFe3C=
R
R+S=
C0 CCFe3C
C
= 0.40 0.022
6.70 0.022 = 0.057
b) Using the lever rule with
the tie line shown
a) Using the RStie line just below the eutectoid
Ca= 0.022 wt% CCFe3C= 6.70 wt% C
Fe3C(cementite)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3C
a+Fe3C
L+Fe3C
C, wt% C
1148C
T(C)
727
C
C0
R S
CFe C3C
Amount of Fe3C in 100 g
= (100 g)WFe3C
= (100 g)(0.057) =5.7 g
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Chapter 10 - 35
Solution to Example Problem (cont.)
c) Using the VXtie line just above the eutectoid and
realizing thatC0= 0.40 wt% C
Ca= 0.022 wt% C
Cpearlite= C= 0.76 wt% C
Fe3C(cementite)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3C
a+Fe3C
L+Fe3C
C, wt% C
1148C
T(C)
727
C
C0
VX
CC
Wpearlite =V
V+X=C0 CC C
= 0.40 0.022
0.76 0.022 = 0.512
Amount of pearlite in 100 g
= (100 g)Wpearlite
= (100 g)(0.512) =51.2 g
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Chapter 10 -
VMSE: Interactive Phase Diagrams
36Change alloy composition
Microstructure, phase compositions, and phase fractions respond interactively
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Chapter 10 - 37
Alloying with Other Elements
Teutectoidchanges:
Adapted from Fig. 10.38,Callister & Rethwisch 4e.(Fig. 10.38 from Edgar C. Bain, Functions of the
Alloying Elements in Steel, American Society for
Metals, 1939, p. 127.)
TEutectoid(C)
wt. % of alloying elements
Ti
Ni
MoSi
W
Cr
Mn
Ceutectoidchanges:
Adapted from Fig. 10.39,Callister & Rethwisch 4e.(Fig. 10.39 from Edgar C. Bain, Functions of the
Alloying Elements in Steel, American Society for
Metals, 1939, p. 127.)
wt. % of alloying elements
Ceutectoid(wt%C
)
Ni
Ti
Cr
Si MnW
Mo
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Chapter 10 - 38
Phase diagramsare useful tools to determine:-- the number and types of phases present,
-- the compositionof each phase,
-- and the weight fraction of each phase
given the temperature and composition of the system. The microstructure of an alloy depends on
-- its composition, and
-- whether or not cooling rate allows for maintenance of
equilibrium.
Important phase diagram phase transformations include
eutectic, eutectoid, and peritectic.
Summary
8/13/2019 Ch 10 Lecture
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Ch t 10 39
Core Problems:
Self-help Problems:
ANNOUNCEMENTS
Reading: